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Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

..I thought With out replacement means we are taking out the balls and not putting it back.

so Total no of ways = 8*7*6*5*4

the no of ways the 1 Red can be drawn = 2 2 Green = 3*2 2 blue = 3*2

so prob = 6*6*2/8*7*6*5*4= 3/280.....

Can any body pls explain the funda of With replacement and Without Replacement.

Let's look at pick two green balls from three green balls. How many ways can you do it? Follow your train of thought, the first pick there's three possibility. The second pick there's only two left, so two possibility. Then you multiply them. However, notice that if the first pick you picked A, and then the second pick you picked B, you would count it as one outcome. If you picked B first, and then A second, you would count it as another outcome. While in fact, (A B) and (B A) is the same outcome, you picked the same two balls. So what you need to do in that case is that you need to divede your number by 2! to get rid of the ordering.

By the same logic for total outcome you have to divide 8*7*6*5*4 by 5!, and you'll get the same result as C(8,5).

Generally, picking n things one by one without replacement can be treated the same as picking n things all together. _________________

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..I thought With out replacement means we are taking out the balls and not putting it back.

so Total no of ways = 8*7*6*5*4

the no of ways the 1 Red can be drawn = 2 2 Green = 3*2 2 blue = 3*2

so prob = 6*6*2/8*7*6*5*4= 3/280.....

Can any body pls explain the funda of With replacement and Without Replacement.

Can somebody can put more light on this solution by probablit method with out using Combinations.

Thanks,Rrsnathan.

Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

A. 8/28B. 9/28C. 10/28D. 10/18E. 11/18

\(P(RGGBB)=\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!2!}=\frac{9}{28}\). We are multiplying by \(\frac{5!}{2!2!}\) because RGGBB case can occur in several ways: RGGBB, GRGBBR, GGRBB, GGBRB, ... (basically it's # of permutations of 5 letters RGGBB, which is \(\frac{5!}{2!2!}\)).

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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